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# If both the roots of the quadratic equation x2 – 2kx + k2 + k – 5 = 0 are less than 5, then k lies in the interval ?

Latika Leekha
7 years ago
The given equation is x2 – 2kx + k2 + k – 5 = 0.
Comparing this equation with the general equation ax2 + bx + c = 0, we get
a = 1, b = -2k, c = k2 + k – 5.
The roots are given by x = [-b ± √ b2 – 4ac] / 2a.
This gives x = [2k ± √ 4k2 – 4k2 – 4k + 20] / 2.
Hence, x = k ± √ (5 – k)
Now, given that roots are less than 5 means x < 5.
k ± √ (5 – k)
First cosider, k+ √(5 – k)< 5
This gives √(5 – k) < (5 – k)
Squaring both sides and solving,
k2 - 9k + 20 < 0.
(k-4)(k-5) < 0
Hence, 4< k < 5.
Now even if we take k – √(5 – k) and proceed in the same way, we shall get the same answer.
Hence, k lies in the intervla (4,5).
Thanks & Regards
Latika Leekha
Sakshi
11 Points
4 years ago

The given equation is x2 – 2kx + k2 + k – 5 = 0.
Comparing this equation with the general equation ax2 + bx + c = 0, we get
a = 1, b = -2k, c = k2 + k – 5.
The roots are given by x = [-b ± √ b2 – 4ac] / 2a.
This gives x = [2k ± √ 4k2 – 4k2 – 4k + 20] / 2.
Hence, x = k ± √ (5 – k)
Now, given that roots are less than 5 means x
k ± √ (5 – k)
± √ (5 – k)
Now, squaring both the sides, we get,
(5-k)2 – 10k
k2 - 9k + 20 > 0
(k-4) (k-5) > 0
so, k lies in the interval (-$\infty$,4) U (5,$\infty$)

Rishi Sharma
one year ago
Dear Student,